Proof of inverse derivative

Petrus

Well-known member
Hello MHB,
I am aware of there is two way, u can use chain rule or defination of derivate. I totaly understand the proof with this type Derivative of Inverse Function but is that a valid proof? How ever our teacher did proof this with derivate defination which I dont understand from my textbook. What is your thought? Any good link that explain this proof with derivate defination

I am aware that we use chain rule and im training for oral exam and I guess I will have to proof this chain rule in this one.

edit: why should $$\displaystyle f'(x) \neq 0$$ should it be $$\displaystyle f'(y) \neq 0$$
Regards,
$$\displaystyle |\pi\rangle$$

Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
re: proof of inverse derivative

Hello MHB,
I am aware of there is two way, u can use chain rule or defination of derivate. I totaly understand the proof with this type Derivative of Inverse Function but is that a valid proof? How ever our teacher did proof this with derivate defination which I dont understand from my textbook. What is your thought? Any good link that explain this proof with derivate defination

I am aware that we use chain rule and im training for oral exam and I guess I will have to proof this chain rule in this one.

edit: why should $$\displaystyle f'(x) \neq 0$$ should it be $$\displaystyle f'(y) \neq 0$$
Regards,
$$\displaystyle |\pi\rangle$$
That proof looks valid to me.

Note that there may be some confusion about x and y, since their meanings are swapped around after the first line.
In the first line x is used as the argument of f, but in the second line and thereafter x is used as the argument of $f^{-1}$ instead (where you might expect y to be the argument).

Petrus

Well-known member
Re: proof of inverse derivative

That proof looks valid to me.

Note that there may be some confusion about x and y, since their meanings are swapped around after the first line.
In the first line x is used as the argument of f, but in the second line and thereafter x is used as the argument of $f^{-1}$ instead (where you might expect y to be the argument).
Thanks for taking your time I like Serena!

PS. Should I be rational or real

Regards,
$$\displaystyle |\pi\rangle$$

Klaas van Aarsen

MHB Seeker
Staff member
Re: proof of inverse derivative

Thanks for taking your time I like Serena!

PS. Should I be rational or real

Regards,
$$\displaystyle |\pi\rangle$$
I think that $$\displaystyle |\pi\rangle$$ is imaginary.